What Can Pi Do for You?

Tomorrow is Pi Day!

And what a Pi Day! 3/14/15 (if you’re in the US, Belize, Micronesia, some parts of Canada, the Philippines, or Swahili-speaking Kenya), best celebrated at 9:26:53, if you’re up by then. Grab a slice of pie, or cake if you really must, and enjoy!

If you don’t have some of your own, download this one!

Pi is great not just because it’s fun to recite digits and eat pastries, but because it serves a very important role in physics. That’s because, often, pi is one of the most “natural” ways to get larger numbers.

Suppose you’re starting with some sort of “natural” theory. Here I don’t mean natural in the technical sense. Instead, I want you to imagine a theory that has very few free parameters, a theory that is almost entirely fixed by mathematics.

Many physicists hope that the world is ultimately described by this sort of theory, but it’s hard to see in the world we live in. There are so many different numbers, from the tiny mass of the electron to the much larger mass of the top quark, that would all have to come from a simple, overarching theory. Often, it’s easier to get these numbers when they’re made out of factors of pi.

Why is pi easy to get?

In general, pi shows up a lot in physics and mathematics, and its appearance can be mysterious the uninitiated, as this joke related by Eugene Wigner in an essay I mentioned a few weeks ago demonstrates:

THERE IS A story about two friends, who were classmates in high school, talking about their jobs. One of them became a statistician and was working on population trends. He showed a reprint to his former classmate. The reprint started, as usual, with the Gaussian distribution and the statistician explained to his former classmate the meaning of the symbols for the actual population, for the average population, and so on. His classmate was a bit incredulous and was not quite sure whether the statistician was pulling his leg. “How can you know that?” was his query. “And what is this symbol here?” “Oh,” said the statistician, “this is pi.” “What is that?” “The ratio of the circumference of the circle to its diameter.” “Well, now you are pushing your joke too far,” said the classmate, “surely the population has nothing to do with the circumference of the circle.”

While it may sound silly, in a sense the population really is connected to the circumference of the circle. That’s because pi isn’t just about circles, pi is about volumes.

Take a bit to check out that link. Not just the area of a circle, but the volume of a sphere, and that of all sorts of higher-dimensional ball-shaped things, is calculated with the value of pi. It’s not just spheres, either: pi appears in the volume of many higher-dimensional shapes.

Why does this matter for physics? Because you don’t need a literal shape to get a volume. Most of the time, there aren’t literal circles and spheres giving you factors of pi…but there are abstract spaces, and they contain circles and spheres. A electric and magnetic fields might not be shaped like circles, but the mathematics that describes them can still make good use of a circular space.

That’s why, when I describe the mathematical formulas I work with, formulas that often produce factors of pi, mathematicians will often ask if they’re the volume of some particular mathematical space. It’s why Nima Arkani-Hamed is trying to understand related formulas by thinking of them as the volume of some new sort of geometrical object.

All this is not to say you should go and plug factors of pi together until you get the physical constants you want. Throw in enough factors of pi and enough other numbers and you can match current observations, sure…but you could also match anything else in the same way. Instead, it’s better to think of pi as an assistant: waiting in the wings, ready to translate a pure mathematical theory into the complicated mess of the real world.

So have a Happy Pi Day, everyone, and be grateful to our favorite transcendental number. The universe would be a much more boring place without it.

7 thoughts on “What Can Pi Do for You?

  1. Tienzen (Jeh-Tween) Gong

    “Throw in enough factors of pi and enough other numbers and you can match current observations, sure…but you could also match anything else in the same way. ”

    Indeed, three π is enough to get very close to (1/137.0359…)


    But, how many pie are needed for the exact (1/137.0359…), as follow?

    Beta = 1/Alpha = 64 ( 1 + first order sharing + sum of the higher order sharing)
    = 64 (1 + 1/Cos A(2) + .00065737 + …)
    = 137.0359 …

    A(2) is the sharing angle, A(2) = 28.743 degree

    The sum of the higher order sharing = 2(1/48)[(1/64) + (1/2)(1/64)^2 + …+(1/n)(1/64)^n +…]
    = .00065737 + …


      1. Tienzen (Jeh-Tween) Gong

        4gravitonsandagradstudent: “Precisely. If you’re willing to dig around, you can match pretty much any number you want with some intuitive-seeming mathematical trick.”

        Agree, 100%.

        Any number can always be reached with zillion different numerological formulas. One the other hand, a single scheme cannot normally reach more than two different numbers.

        Planck data (dark energy = 69.2; dark matter = 25.8; and visible matter = 4.82) consists of three seemingly unrelated numbers. The following is a single scheme.

        Among 48 fermions (quarks and leptons), only 7 of them are visible.

        So, the d/v (dark/visible ratio) = [41 (100 – W) % / 7]
        When, W = 9 % (according to the AMS2 data), d/v = 5.33

        In this scheme, the space, time and mass form an iceberg model.

        Space = X
        Time = Y
        Total mass (universe) = Z
        And X = Y = Z

        In an iceberg model (ice, ocean, sky), Z is ice while the (X + Y) is the ocean and sky, the energy ocean (or the dark energy). Yet, the ice (Z) will melt into the ocean (X + Y) with a ratio W.

        When W = 9%,
        [(Z – V) x (100 – W) %] /5.33 = V, V is visible mass of this universe.
        [(33.33 –V) x .91]/5.33 = V
        V= 5.69048 / 1.17073 = 4.86 (while the Planck data is 4.82),
        D (Dark mass) = [(Z – 4.86) x (100 – W) %] = [(33.33 -4.86) x .91] = 25.90 (while the Planck data = 25.8)

        So, the total dark energy = (X + Y) + [(Z – 4.86) x W %)] = 66.66 + (28.47 x 0.09) = 69.22 (while the Planck data is 69.2)

        Except the ‘W’ is a free parameter (testable), the above calculation is purely theoretical, and it marches the data to an amazing degree.

        In fact, it is this same scheme giving rise to the Alpha equation and {246/2 + 2.46 = 125.46 Gev} boson.

        The same scheme gives rise to {force F (dark energy) = ħ/ (delta S x delta T) to (delta P x delta S > =ħ}


        1. 4gravitonsandagradstudent Post author

          Considering that there’s at least one other parameter in your comment itself. (“Among 48 fermions (quarks and leptons), only 7 of them are visible.”), let alone how many possible “iceberg models” there are, claiming that W is the only free parameter in your setup is remarkably disingenuous.



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